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Related papers: Tangent power sums and their applications

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Let $p$ be an odd prime, and let $a$ be an integer not divisible by $p$. When $m$ is a positive integer with $p\equiv1\pmod{2m}$ and $2$ is an $m$th power residue modulo $p$, we determine the value of the product $\prod_{k\in…

Number Theory · Mathematics 2023-08-25 Zhi-Wei Sun

Via symbolic computation we deduce 97 new type series for powers of $\pi$ related to Ramanujan-type series. Here are three typical examples: $$\sum_{k=0}^\infty \frac{P(k) \binom{2k}k\binom{3k}k…

Number Theory · Mathematics 2020-07-17 Zhi-Wei Sun

We prove congruences, modulo a power of a prime p, for certain finite sums involving central binomial coefficients $\binom{2k}{k}$.

Number Theory · Mathematics 2013-10-09 Sandro Mattarei , Roberto Tauraso

The following congruence for power sums, $S_n(p)$, is well known and has many applications: $1^n+2^n +\dots +p^n \equiv\begin{cases} -1 \text{ mod } p, & \text{ if } \ p-1 \ | \ n; 0 \text{ mod } p, & \text{ if } \ p-1 \ \not| \ n,…

Number Theory · Mathematics 2018-01-08 Nicholas J. Newsome , Maria S. Nogin , Adnan H. Sabuwala

We present a new and elegant integral approach to computing the Gardner-Fisher trigonometric power sum, which is given by $$ S_{m,v}=\left(\frac \pi{2m}\right)^{2v}\sum_{k=1}^{m-1}\cos^{-2v}\left(\frac{k\pi}{2m}\right)\, , $$ We present a…

Number Theory · Mathematics 2016-03-14 Carlos M. da Fonseca , M. Lawrence Glasser , Victor Kowalenko

We show how to find series expansions for $\pi$ of the form $\pi=\sum_{n=0}^\infty {S(n)}\big/{\binom{mn}{pn}a^n}$, where S(n) is some polynomial in $n$ (depending on $m,p,a$). We prove that there exist such expansions for $m=8k$, $p=4k$,…

Number Theory · Mathematics 2007-05-23 Gert Almkvist , Christian Krattenthaler , Joakim Petersson

Let f be a sum of exponentials of the form exp(2 pi i N x), where the N are distinct integers. We call f an idempotent trigonometric polynomial (because the convolution of f with itself is f) or, simply, an idempotent. We show that for…

Classical Analysis and ODEs · Mathematics 2007-05-23 Bruce Anderson , J. Marshall Ash , Roger Jones , Daniel G. Rider , Bahman Saffari

We show that if there exists an integer subject to some congruence conditions that cannot be written as the sum of the norm of an ideal in $\mathbb{Z}[\exp(2\pi i/2^k)]$ and at most $k$ powers of $2$, $k\geq 3$, then there are infinitely…

Number Theory · Mathematics 2016-10-19 Timothy Foo

We provide a new, simple general proof of the formulas giving the infinite sums $\sigma(p,\nu)$ of the inverse even powers $2p$ of the zeros $\xi_{\nu k}$ of the regular Bessel functions $J_{\nu}(\xi)$, as functions of $\nu$. We also give…

Mathematical Physics · Physics 2014-02-14 Jorge L. deLyra

In this paper we investigate a certain category of cotangent sums and more specifically the sum $$\sum_{m=1}^{b-1}\cot\left(\frac{\pi m}{b}\right)\sin^{3}\left(2\pi m\frac{a}{b}\right)\:$$ and associate the distribution of its values to a…

Classical Analysis and ODEs · Mathematics 2017-09-21 Michael Th. Rassias

We present several congruences modulo a power of prime $p$ concerning sums of the following type $\sum_{k=1}^{p-1}{m^k\over k^r}{2k\choose k}^{-1}$ which reveal some interesting connections with the analogous infinite series.

Number Theory · Mathematics 2009-12-20 Roberto Tauraso

For each 1 < p < infinity, there exists a positive constant c_p, depending only on p, such that the following holds. Let (d_k), (e_k) be real-valued martingale difference sequences. If for for all bounded nonnegative predictable sequences…

Probability · Mathematics 2007-05-23 Stephen Montgomery-Smith , Shih-Chi Shen

We give an exact coefficients formula of any infinite product of power series with constant term equal to $1$, by using structures from partitions of integers and permutation groups. This is an universal theorem for various of Binomial-type…

Combinatorics · Mathematics 2024-11-05 Kui-Yo Chen , Zhong-Tang Wu

Let $n$ be a non-negative integer and put $p_{n}(x)=\prod_{i=0}^{n}(x+i)$. In the first part of the paper, for given $n$, we study the existence of integer solutions of the Diophantine equation $$ y^m=p_{n}(x)+\sum_{i=1}^{k}p_{a_{i}}(x), $$…

Number Theory · Mathematics 2018-09-13 Szabolcs Tengely , Maciej Ulas

We study divisibility properties of certain sums and alternating sums involving binomial coefficients and powers of integers. For example, we prove that for all positive integers $n_1,..., n_m$, $n_{m+1}=n_1$, and any nonnegative integer…

Number Theory · Mathematics 2012-04-10 Victor J. W. Guo , Jiang Zeng

Sum of powers 1^p+...+n^p, with n and p being natural numbers and n>=1, can be expressed as a polynomial function of n of degree p+1. Such representations are often called Faulhaber formulae. A simple recursive algorithm for computing…

Discrete Mathematics · Computer Science 2009-03-26 M. Torabi Dashti

We revisit Euler's partition function recurrence, which asserts, for integers $n\geq 1,$ that $$ p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+\dots = \sum_{k\in \mathbb{Z}\setminus \{0\}} (-1)^{k+1} p(n-\omega(k)), $$ where $\omega(m):=(3m^2+m)/2$ is…

Number Theory · Mathematics 2025-04-22 Kevin Gomez , Ken Ono , Hasan Saad , Ajit Singh

We give a characterization of all pairs $(k,n)$ of positive integers for which the ratio $$ \frac{1^k-2^k+3^k-\dots+(-1)^{n+1} n^k}{1^k-2^k+3^k-\dots+(-1)^{n}(n-1)^k} $$ of two consecutive alternating power sums is an integer.

Number Theory · Mathematics 2019-07-16 Ioulia N. Baoulina

In this manuscript we provide a new polynomial pattern. This pattern allows to find a polynomial expansion of the form \[x^{2m+1} = \sum_{k=1}^{x}\sum_{r=0}^{m} \mathbf{A}_{m,r} k^r (x-k)^r,\] where $x,m\in\mathbb{N}$ and $\mathbf{A}_{m,r}$…

General Mathematics · Mathematics 2022-11-01 Petro Kolosov

An integer of the form $p_m(x)= \frac{(m-2)x^2-(m-4)x}{2} \ (m\ge 3)$, for some integer $x$ is called a generalized polygonal number of order $m$. A ternary sum $\Phi_{i,j,k}^{a,b,c}(x,y,z)=ap_{i+2}(x)+bp_{j+2}(y)+cp_{k+2}(z)$ of…

Number Theory · Mathematics 2021-02-10 Jangwon Ju , Byeong-Kweon Oh , Bangnam Seo
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